Product Description:
610kV Oilimmersed DistributionTransformers
1. Introduction
610kV oilimmersed distributiontransformers with capacity 303150kVA are suitable for 610kV distributionnetwork system. Product complies with GB/T64512008 standards.
Type 9, 11 and 13 are classified byloss standard. Two types of oil tank are provided, they are sealed corrugatedradiating and sheet radiator. Voltage can be adjusted with load through switchinstalled.
2. Characteristics
Fully sealed corrugated tank has corrugatedsheet for transformer oil natural cooling. There is no oil storage cabinetneeded, so the oil is completely isolated from air. This design slows down theaging of oil, eliminates maintenance and extends the transformer life.
3. Technical parameters
Technical parameters of S9Mcorrugated tank transformers
Code  Rated capacity (kVA)  rated voltage(kV)  Loss (xW)  Short Circuit impedance (%)  No Load Current (%)  Connection Symbol  Weight (kg)  Dimension L*W*H (mm)  Gauge mm 

HV  LV  No Load  Load 

Body  Oil  Total 

S9M100  100  10 6.3 6  6.3 6  290  1500/1580  4.0  1.8  Yyn0 Dyn11  285  105  515  800x680x1030  550/550 

S9M125  125  340  1800/1890  1.7  355  110  625  920x700x1060  550/550 

S9M160  160  400  2200/2310  1.6  420  130  740  970x730x1070  550/550 

S9M200  200  480  2600/2730  1.5  475  160  825  1020x740x1085  550/550 

S9M315  315  670  3650/3830  1.4  710  205  1200  1350x740x1115  550/550 

S9M500  500  960  5150/5410  1.2  925  285  1575  1470x795x1310  660/660 

S9M630  630  1200  6200  4.5  1.1  1085  320  1835  1540x830x1360  660/660 

S9M800  800  1400  7500  1.0  1375  435  2340  1630x920x1430  820/820 

S9M1000  1000  1700  10300  1.0  1480  485  2600  1755x1035x1450  820/820 

S9M1250  1250  1950  12000  0.9  1760  575  3135  1870x1120x1540  820/820 

S9M1600  1600  2400  14500  0.8  2125  700  3795  1970x1170x1645  820/820 

Technicalparameters for S11M corrugated tank transformers
Code  Rated capacity (kVA)  rated voltage(kV)  Loss (xW)  Short Circuit impedance (%)  No Load Current (%)  Connection Symbol  Weight (kg)  Dimension L*W*H (mm)  Gauge mm 
HV  LV  No Load  Load     Body  Oil  Total   
S11M160/10  160  10 6.3 6  0.4 6.3 6  280  2200/2310  4.0  1.6  Yyn0 Dyn11  490  100  795  1135x713x180  550/550 
S11M200/10  200  340  2600/2730  1.5  540  175  880  1168x738x1200  550/550 
S11M250/10  250  400  3050/3200  1.4  665  200  1025  1240x780x1240  550/550 
S11M315/10  315  480  3650/3830  1.3  730  220  1155  1300x835x1265  550/550 
S11M400/10  400  570  4300/4520  1.2  895  300  1390  1390x905x1310  550/550 
S11M500/10  500  680  5150/5410  1.1  1025  360  1650  1470x965x1365  550/550 
S11M630/10  630  0.4  810  6200  4.5 5.5  1.0  1650  415  2260  1575x1010x1430  660/660 
S11M800/10  800  980  7500  1.0  1510  470  2550  1685x940x1545  820/820 
S11M1000/10  1000  1150  10300  0.9  Dyn11  1640  650  3070  2180x1075x1655  820/820 
S11M1250/10  1250  1360  12800  0.8  2010  820  3700  2310x1310x1715  820/820 
S11M1600/10  1600  1640  14500  0.8  2420  990  4470  2460x1514x1920  820/820 
S11M2000/10  2000  2240  16830  0.8  2860  1120  5050  2782x1600x2040  820/820 
S11M2500/10  2500  2640  19550  0.8  3195  1155  5912  2500x2060x2010  820/820 
Parameters for 610kV SZ11 OLTC Transformers
Code  Rated capacity (kVA)  Rated voltage (kV)   Loss(xW)  Short Circuit impedance (%)  No Load Current (%)  Connection Type Symbol  Weight (kg)  Dimension L*W*H(mm)  Gauge mm） 
HV  LV  No Load  Load  Body  Oil  Total 
S11M200/10  200  10 6.3 6  0.4  ±4x2.5%  0.48  3.06  4.0  1.5  Yyn0 Dyn11  531  241  1058  1510x820x1460  550/550 
S11M250/10  250  0.56  3.60  1.4  630  350  1400  1740x1020x1440  550/550 
S11M315/10  315  0.67  4.32  1.4  860  280  1630  1790x1050x1570  550/550 
S11M400/10  400  0.80  5.22  1.3  910  400  1740  1830x1120x1630  660/660 
S11M500/10  500  0.96  5.21  1.2  1080  460  2068  1900x1230x1780  660/660 
S11M630/10  630  1.20  7.65  1.1  1396  611  2661  2010x1320x1930  660/660 
S11M800/10  800  1.40  9.36  1.0  1650  780  3220  2280x1370x2220  820/820 
S11M1000/10  1000  1.70  10.98  1.0  2083  843  4240  2170x1160x2320  820/820 
S11M1250/10  1250  1.95  13.05  0.9  2390  1100  4950  2510x1310x2630  820/820 
S11M1600/10  1600  2.40  15.00  0.8  2900  1065  5235  2570x1382x2650  820/820 
Note: The above parameter before slash is forYyn0 connection type and after slash is for Dyn11 connection type.
 Q:Two Transformers Parallel Operation
 If the transformer voltage ratio is equal (equivalent to the equivalent of the induced potential), the impedance voltage (equivalent to the transformer internal resistance) are equal, then they output the current is equal, that is, the load is evenly distributed. The total capacity is the sum of their capacity. Otherwise, the impedance of the output voltage of the small current, its load rate is relatively high, when its load current reaches full load, the impedance voltage has not reached a full load, if the load at this time for the total Capacity, then the total capacity is less than the sum of the two capacity. This is a strict argument. In fact, the two transformers of the impedance voltage difference is very small, the load distribution is also very small difference, in addition, the socalled transformer is a little overload, nor is it so strict, so the actual total capacity and the difference between the two capacity Not big.
In order to prevent the load distribution is too uneven, the capacity of these two transformers should be as close as possible, should not exceed 3: 1, because the capacity of the large impedance voltage is small, it has reached full load, and that small capacity is still in the owe Load, its role is not fully play out, it is not the significance of the parallel.
The above is the answer 2009108 02:31.
At the end of the said, "the large capacity of the impedance voltage is small, it has reached full load, and that small capacity is still in the obvious underload, its role is not fully play out", wrong, apologize. Should be large capacity transformer impedance voltage, and small capacity impedance voltage is small, the load rate is high, the output current to reach full load, resulting in a larger capacity of the transformer can not put its big capacity advantage fully play out, this is Very unfavorable.
 Q:If I have a 600 watt 12 volt transformer plugged into a 120 volt circuit, what 120v. amperage is needed?
 A circular saw is an AC circuit. If you were using a DC circuit, you would be using a 1200 watt tool since all of the electricity (subtracting heat loss of course) is used to run the tool. In an AC circuit, the AC is 'alternating current', a sine wave. Only a portion of the sine wave is used for real work. The rest is used to establish and collapse magnetic fields. This term is known as 'Root Mean Square' or RMS and has to be taken into account for the problem. The value of RMS in a typical 120 volt, 60 Hz circuit is .707 or 70.7% of usable energy in the sine wave. So this leads us to the question, how many watts? Answer(120V * 10A).707 848.4W. The next one is easy, they are already watt rated so there's no conversion. 1000 Watts * 8 hrs 8KWH. 8 cents * 8 KWH 64 cents. Hope this helps.
 Q:Transformer 1250KVA Dimensions
 This is not fixed, you list the conventional products, unconventional products can be arbitrarily selected according to customer requirements. There are 300kv, 1000KV, 1100KV, 1150KV and so on
 Q:i have a 12  220v transformer, the primary makings are 12v, 0v. 12v is this a 24v center tapp transformer?
 confident. you will incur larger losses than if the transformer became particularly designed for 220 V, inspite of the undeniable fact that it is going to in all possibility be insignificant. be conscious, inspite of the undeniable fact that, that attempting the different (making use of a transformer designed for 220 V in a 440 V circuit) might _not_ paintings, because of the fact the conventional might overheat.
 Q:Why the computer water does not use transformer oil
 In terms of cooling medium, the cooling effect of water is very good, it is used in the weak is the most used to achieve the purpose of cooling and the lowest operating costs.
 Q:design a quarter wave transformer to match a 73 ohms antenna to 300 ohms transmission line fed by a 100MHz FM?
 Transformer impedance required: sqrt (73 x 300) 147.986 (say 148) ohms. The 1/4 wavelength in air is (75/100) 0.75 metres. You have not stated what you are using for the transformer, so you will need to allow for its velocity factor. But the transformer could be 0.75 metres of parallel rod spaced as required for the impedance. I am also assuming the antenna is balanced (sounds like a folded dipole).
 Q:Question on transformers?
 Since there is no evidence to indicate any inefficiency of this transformer, you have to assume that it is 100% efficient, with input power equaling output power. Input power Vin * Iin Output power Vout*Iout Equate: Vin*Iin Vout*Iout Solve for Iin, our unknown: Iin Iout * Vout/Vin Data: Vout 110 V Vin 230 V Iout 0.15 A Result: Iin 0.072 As for the ratio of loops, the number of turns in each coil is proportional to the voltage across each coil. Thus, the turns ratio equals the voltage ratio of 110/230 0.478. That is to say, there are 2.1 turns in the primary coil for every individual turn there is in the secondary coil. The primary coil is the coil plugged in to the line voltage of 230 V. The secondary coil is the coil in to which the load plugs in, at load voltage of 110 V.
 Q:What is transformer saturation?
 Transformer Saturation
 Q:Homework on transformers?
 The turns ratio is 4:1 of the primary. The power supplied to the primary is 400W. So regardless of anything else, in a ideal transformer, 400W is available on the secondary. But voltage will be 4x higher, current 4x smaller, than the voltage and current in the primary. Disregarding all that, if there is 400W of power flowing in the primary, then 400W will be being dissipated in the 20 ohm resistor. Watts in watts out, but secondary load (resistance) controls the wattage. There is no way to tell the output voltage and current, without knowing the input voltage and current. But whatever they are, the product of voltage x current will be 400W, in this instance.
 Q:Proteus transformer how to adjust the parameters, the 220V into 14V output.
 Transformer parameter setting method is:
1, the calculation: According to the ideal transformer primary turns ratio ratio n1 / n2 = V1 / V2 = 220/14 ≈ 15.7,
Let the primary winding sense L1 = 1H, then the secondary winding inductance L2 = 1 / (15.7 ^ 2) ≈ 0.00406H,
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